What is the minimum value of the radius ratio | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) In an ionic crystal,the stability of the lattice depends on the radius ratio ${r_ + }/{r_ - }$. For an octahedral void (coordination number $6$),w

Vedclass · Accepted Answer

$0.414$

Vedclass · Answer

(A) In an ionic crystal,the stability of the lattice depends on the radius ratio ${r_ + }/{r_ - }$. For an octahedral void (coordination number $6$),which is the structure of $NaCl$,the limiting radius ratio is calculated as follows: In an octahedral void,the cation is surrounded by $6$ anions. For the condition where the cation just touches the anions and the anions do not touch each other,the geometry leads to the relation: ${r_ + } + {r_ - } = \sqrt{2} {r_ - }$. Rearranging this,we get ${r_ + } = (\sqrt{2} - 1) {r_ - }$. Thus,${r_ + }/{r_ - } = \sqrt{2} - 1 \approx 1.414 - 1 = 0.414$. Therefore,the minimum radius ratio for an octahedral arrangement is $0.414$.

What is the minimum value of the radius ratio ${r_ + }/{r_ - }$ for $NaCl$ to ensure that anion-anion contact does not occur?

Similar Questions

If the pressure on a $NaCl$ structure is increased,then its coordination number will

$A$ compound is formed by elements $A$ and $B$. Element $A$ occupies the corners of the cube,and element $B$ occupies the face centers. What is the formula of the compound?

$A$ cubic unit cell contains manganese ions at the corners and fluoride ions at the centres of each edge. The coordination number of the manganese ion is-

Explain the calculation of the number of atoms in a primitive cubic unit cell.

The tetrahedral voids formed by $ccp$ arrangement of $Cl^{-}$ ions in rock salt structure are

Vedclass Test Series

Exam Paper Generator

Online Exam Module